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4 years ago

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A rectangular squeeze film bearing will be used to absorb intermittent loads. The bearing is 500mm wide. The initial gap between

the bearing plates is 2mm. The applied load is 50N. Oil viscosity 1000 mPa-s. a) If the bearing gap should change to 1mm in 0.5s, how long will you make the bearing? b) Using your calculated length, how long will it take to move if the viscosity is changed to 300 mPa-s?

1 answer:

Anna71 [15]4 years ago

6 0

Answer:

300 yeet

Explanation:

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A solid sphere is rolling on a surface. what fraction of its total kinetic energy is in the form of rotational kinetic energy ab

drek231 [11]

<span>Total KE = KE (rotational) + KE (translational) Moment of inertia of sphere is I = (2/5)mr^2 So KE (rotational) = (1/2) x I x w^2 = (1/2) x (2/5)mr^2 x w^2 = (1/5) x m x r^2 x w^2 KE (translational) = (1/2) x m x v^2 = (1/2) x m x (rw)^2 = (1/2) x m x r^2 x w^2 Hence KE = (1/5) x m x r^2 x w^2 + (1/2) x m x r^2 x w^2 = m x r^2 x w^2 ((1/5) + (1/2)) KE = (7/10) m x r^2 x w^2 Calculating the fraction of rotational kinetic energy to total kinetic energy, = rotational kinetic energy / total kinetic energy = (1/5) x m x r^2 x w^2 / (7/10) m x r^2 x w^2 = (1/5) / (7/10) = 2 / 7 The answer is 2 / 7</span>

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3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

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Answer:

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Explanation:

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An example of a Destructive force is the formation of mountains True or False?

nata0808 [166]

Hey there!

<span>An example of a Destructive force is the formation of mountains.

This statement is false, an example of a destructive force would not be the formation of mountains.

Hope this helps you.
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